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Let's say I can measure something with a know standard deviation of $\sigma$. Let's say I measure it $n$ times and take the mean of the $n$ measurements to determine a more accurate mean value. What is the new $\sigma$ for my mean measurement? I can intuitively know that it'll be smaller than the original $\sigma$ but still greater than $0$, but how does one calculate the new $\sigma$ value?

Bonus question (but much harder I suspect):

What if still have the $n$ measurements, but they all have slightly different $\sigma$. How do I now compute/derive the new $\sigma$ (given that the $n$ measurements all had slightly different $\sigma$). (In case it helps, $n$ will typically be ~$40$, but it can be as little as $10$ or as much as $500$; also, the different $\sigma$ do vary, but not by much: the smallest $\sigma$ and the biggest $\sigma$ will differ by at most ~$2\times$)

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    $\begingroup$ Are you talking about standard deviation or standard error? Those terms are related but not synonymous, and the answer to your questions depends on what you mean. $\endgroup$ – Dave Oct 26 '19 at 15:48
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    $\begingroup$ Sampling from a normal population with mean $\mu$ and SD $\sigma,$ the best estimate of the population mean $\mu$ is the sample mean $\bar X,$ The std error of the mean is $SD(\bar X) = \sigma/\sqrt{n}.$ The (estimated) std error of the mean is $S/\sqrt{n},$ where $S$ is the sample SD. Either way, the denominator $\sqrt{n}$ implies that the std error gets smaller with increasing $n.$ If $\sigma$ is known a 95% confidence interval for $\mu$ is of the form $\bar X \pm 1.96\frac{\sigma}{\sqrt{n}}.$ // Your question is not clear. In addition to the standard facts above, for what are you asking? $\endgroup$ – BruceET Oct 26 '19 at 18:36
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    $\begingroup$ If the SD is known, I presume you are asking "how fast does the sample statistic for SD converge?". It is both surprisingly slow, and very hard to find - even on the internet. If that is the question, I'll try to hunt up a few good ferences I may have bookmarked. $\endgroup$ – eSurfsnake Oct 27 '19 at 3:58
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Let's consider the "bonus question" first, because the first question is just a simple special case.

Standard deviations are square roots of variances. Square roots are difficult to work with, so restate your problem in terms of measurement variances and compute the standard deviation of the mean only once you have a solution for its variance.

For independent measurements, variances add. This means that the variance of the sum of measurements $S=X_1 + X_2 + \cdots + X_n$ is the sum of the measurement variances $\sigma_i^2,$ whence

$$\operatorname{Var}(S) = \sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2.$$

Variances scale quadratically. This means that the variance of a multiple $\lambda$ of any random variable $S$ is $\lambda^2$ times the variance of $S.$ Applying this to $\lambda=1/n$ gives the variance of the mean of your measurements. Its square root is the standard deviation:

$$\operatorname{sd}\left(\frac{S}{n}\right) = \sqrt{\operatorname{Var}\left(\frac{S}{n}\right)} = \sqrt{\left(\frac{1}{n}\right)^2\left(\sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2\right)} = \frac{1}{n}\sqrt{\sigma_1^2 + \sigma_2^2 + \cdots + \sigma_n^2}.$$


In the case of $n$ measurements of the same quantity, where every $\sigma_i$ is the same value $\sigma,$ this shows the standard deviation of the average of those measurements is

$$\frac{1}{n}\sqrt{\sigma^2 + \sigma^2 + \cdots + \sigma^2} = \frac{1}{n}\sqrt{n\sigma^2} = \frac{\sigma}{\sqrt{n}},$$

answering the first question.

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